# Eigenvalues of an operator and its adjoint

I would like to verify the proof given in an answer #2 form here. The claim to prove was that $$\lambda$$ is an eigenvalue of $$T$$ if and only if $$\overline{\lambda}$$ is eigenvalue of the adjoint operator $$T^*$$ (I assume finite dimensional inner product space). Since:

$$\langle T^*u, v \rangle = \overline{\langle Tv, u \rangle } = \overline{\langle\lambda v, u \rangle } = \overline{\overline{\lambda}\langle v,u\rangle } = \lambda\langle u,v\rangle = \langle\overline{\lambda} u,v \rangle \space\quad (1)$$

and similarly

$$\langle Tv, u \rangle = \overline{\langle T^*u,v\rangle} = \overline{\langle \overline{\lambda}u,v \rangle} = \overline{\lambda \langle u,v \rangle} = \overline{\lambda}\langle v,u\rangle = \langle \lambda v, u\rangle \quad(2)$$

The answer seems (to me) to imply that since $$\langle T^*u, v \rangle = \langle\overline{\lambda}u,v \rangle$$, this implies $$T^*u = \overline{\lambda}u \$$ from $$(1)$$. But isn't it the case that this is true only if it holds for all vectors $$v$$, whereas $$v$$ in $$(1)$$ is an eigenvector? And similarly isn't it the case that $$\langle Tv, u \rangle = \langle \lambda v,u \rangle$$ implies $$Tv=\lambda v$$ only if it holds for all vectors $$u$$?